Solve the matrix equation Ax = 0. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set x1 = t and solve for x2 and x3 in terms of t.)

A =
3 -1 -1
1 -3 3
0 =
0
0
Find: (x1, x2, x3) =-

Step-by-step explanation:

Converting the matrix to a set of simultaneous equations we get :

3x1- x2- x3 = 0

x1- 3x2 + 3x3 = 0

Clearly there will be an infinite number of solutions as there is less number of equations than the number of unknowns.

solve for t from the second equation : t = 3x2- 3x3

Substitute (3x2- 3x3) for t in the first equation : 3(3x2-3x3) - x2 - x3 = 0

Substitute (5x3/4) for x2 in the equation for t we get : t=3(5x3/4) - 3x3

Now substitute (4t/3) for x3 in the equation for x2 we get : x2 = 5/4(4t/3)

Hence (x1, x2, x3) = ( t, 5t/3 , 4t/3)

a² + b²= c²

24² + 7² = c²

c= 25

1

step-by-step explanation:

it's the only multiple of both and the least